Why is 0x20 subtracted from the stack pointer in the prolog of this function code?

Question

void main(){ int c; c = function(1, 2); } int function(int a, int b){ char buf[10]; a = a+b; return a; } 

Build Code:

  main: 08048394: push %ebp 08048395: mov %esp,%ebp 08048397: and $0xfffffff0,%esp **0804839a: sub $0x20,%esp <-----------------------???????** 0804839d: movl $0x2,0x4(%esp) 080483a5: movl $0x1,(%esp) 080483ac: call 0x80483b7 <function> 080483b1: mov %eax,0x1c(%esp) 080483b5: leave 080483b6: ret function: 080483b7: push %ebp 080483b8: mov %esp,%ebp 080483ba: sub $0x10,%esp 080483bd: mov 0xc(%ebp),%eax 080483c0: add %eax,0x8(%ebp) 080483c3: mov 0x8(%ebp),%eax 080483c6: leave 080483c7: ret 

I know, aligned on 16 bytes,
but in the expression main (), int c(=4 byte) + 1(4byte) + 2(4byte) in function(1 ,2) .

therefore, the sum of this is 12 bytes. but for a memory-aligned address, I evaluate 16 bytes.

 (sub 0x10, %esp) 

why sub 0x20, %esp ?

Answer 1

Consider this function:

 void main(){ int c, d, e, f; c = function(1, 2, 3, 4); d =1; e = 2; f = 3; } 

However, this will allocate 0x20 space.

But if you add another local parameter or function parameter, it will immediately allocate 0x30 space.

Now consider when there is nothing in the main function, but only one statement:

 int c = 1; 

Then in this case 0x10 space will be allocated.

Do you see the template here? First, the system allocates space for the local variable. Then it will allocate space for function parameters. Allocated space is aligned to 0x10.

That is why you see 0x20. 0x10 for local variables, and another 0x10 for function parameters.